Of course since f is surjective (injective), then each fi is surjective (injective). Thus each fi is an isomorphism, because each Gi is Hopfian (co-Hopfian). Therefore f is an isomorphism, because

f-1=⊕i∈Ifi-1.□

,,⇐” Fix j∈I and assume that fj:Gj→Gj is a surjective (injective) homomorphism. For i∈I such that i≠j define fi:Gi→Gi to be any automorphism of Gi. Then

⊕i∈Ifi:⊕i∈IGi→⊕i∈IGi

is a surjective (injective) group homomorphism. Since ⊕i∈IGi is Hopfian (co-Hopfian) then ⊕i∈Ifi is an isomorphism. Thus each fi is an isomorphism. In particular fj is an isomorphism, which completes the proof. □

Example. Let 𝒫={p∈ℕ|p⁢ is prime} and 𝒫0 be any subset of 𝒫. Then

⊕p∈P0ℤp

is both Hopfian and co-Hopfian.

Proof. It is easy to see that {ℤp}p∈𝒫 is full, so {ℤp}p∈𝒫0 is also full. Moreover for any p∈P0 the group ℤp is finite, so both Hopfian and co-Hopfian. Therefore (due to proposition)